Zener diodes vs voltage divider

[level6]

I'm looking over the datasheet of an IC and it has a control voltage pin that has an input range of 0 to +2V. I'm also looking over the schematics of some designs using this chip that use voltage dividers to limit the input voltage. Wouldn't a 2V Zener be better? If you're using a voltage divider there is an expectation of incoming voltage range. With a Zener diode, that would not be a concern.

[wasedadoc]

Have you tried to find a 2 Volt Zener?

[level6]

Sure. Mouser has a large selection.

[dietert1]

Low voltage zeners are bad as they take current in reverse direction even below zener voltage. I remember trying to use a zener as protection of the 3.3 V supply voltage for a low power MCU. At 3.3 V the 3.6 V zeners would take about 1 mA. We unsoldered them from all boards.
Anyway, the lowest one i know is made of three diodes in forward direction to get 2.1 V.

Regards, Dieter

[level6]

Would that be the reverse leakage current? The part I'm looking at has a max value of 120uA.

https://www.mouser.com/ProductDetail/Vishay-Semiconductors/GDZ2V0B-G3-08?qs=S660mHfF5o0CMOi1bEdkVQ%3D%3D

[SeanB]

Forward leakage current, look at the curve for Vf versus current, and you will see there is considerable current at low voltage, and the zener voltage is specified at a certain current. Even worse this current varies with both temperature and time, and also with stray light hitting the junction. Low voltage zener diodes are very poor, generally the application has them run at a specified current, typically 5 to 50mA, as per datasheet, and you only will get the voltage there, higher current higher voltage, and you can also look at slope resistance to predict the voltage change.

You want to use it to protect you run it from a higher voltage via a resistor, and the use a low leakage diode to switch current from input overload into it, you typically use the BC junction of a bipolar transistor, as they have low leakage, and normally a 30V reverse voltage. Need 2, one for positive peaks, and one from negative, or use another transistor to provide that clamp, and accept the increased power dissipation in the zener clamp.

[bdunham7]

Would that be the reverse leakage current? The part I'm looking at has a max value of 120uA.

https://www.mouser.com/ProductDetail/Vishay-Semiconductors/GDZ2V0B-G3-08?qs=S660mHfF5o0CMOi1bEdkVQ%3D%3D

That datasheet doesn't have a reverse I-V curve which is what you need to understand the problem with low voltage zeners.  Take a look at a datasheet for a different line of zeners that does show this relationship, compare the response of the low-voltage zener in Fig. 6 with the normal and high voltage ones in Fig. 7 and 8.

https://www.mouser.com/datasheet/2/916/BZX38450_Q_SER-2580608.pdf

[DavidAlfa]

First: What IC are you taking about?

[level6]

First: What IC are you taking about?

AS3360

https://alfarzpp.lv/eng/sc/AS3360.pdf

[DavidAlfa]

The source providing the voltage should be designed to ensure it doesn't output more than 2V in any case, but if you want additional protection with the zener it should work just fine.

[level6]

The source providing the voltage should be designed to ensure it doesn't output more than 2V in any case, but if you want additional protection with the zener it should work just fine.

There would be somewhat limited control over the incoming voltage range. The expected external input range is 0-5V, but since the source is a cable being inserted into a jack, I'm thinking I could use a voltage divider to limit to the expected range and a Zener to handle any over-voltage. That way I could avoid the worst of the current leakage.

[Manul]

Zener will have too soft knee. It will not protect well and it will cause non-linear loading near 2V which might cause error in your control voltage which will be hard to calibrate out.

I don't know which IC it is, but it might be that:

1. 2V is not exactly absolute maximum and you can go above that without issues, or
2. There is internal clamping above 2V which means that with limited current it is safe to apply higher voltages.

So it very well might be that all you need is a resistor divider (which also does a job of current limitting).

[level6]

Zener will have too soft knee. It will not protect well and it will cause non-linear loading near 2V which might cause error in your control voltage which will be hard to calibrate out.

I don't know which IC it is, but it might be that:

1. 2V is not exactly absolute maximum and you can go above that without issues, or
2. There is internal clamping above 2V which means that with limited current it is safe to apply higher voltages.

So it very well might be that all you need is a resistor divider (which also does a job of current limitting).

Good point about the non-linear response of the zener.

The absolute maxinum on the pin input is 2.5V, so I can't go much higher than 2V.

[bdunham7]

The absolute maxinum on the pin input is 2.5V, so I can't go much higher than 2V.

OK, so how high do you need the signal to go without any excess leakage from whatever protection method you choose?  And how much added input impedance from an series input resistor can you tolerate?

[level6]

The absolute maxinum on the pin input is 2.5V, so I can't go much higher than 2V.

OK, so how high do you need the signal to go without any excess leakage from whatever protection method you choose?  And how much added input impedance from an series input resistor can you tolerate?

As high as +-10V.

My plan is to have the external input go into an opamp to scale down the voltage to an expected value, then re-inverted on another opamp voltage follower to get the correct polarity. I need to handle the case where a higher voltage than expected is externally supplied.

[Manul]

The absolute maxinum on the pin input is 2.5V, so I can't go much higher than 2V.

Ok, and what happens above 2.5V? Most modern ICs integrate some sort of clamping. For example, Op Amps often include differential mode clamps, some of them safely sink many milliamps. I bet there is something like that in your IC. In that case simply limiting current is a very good protection.

My plan is to have the external input go into an opamp to scale down the voltage to an expected value, then re-inverted on another opamp voltage follower to get the correct polarity. I need to handle the case where a higher voltage than expected is externally supplied.

It's up to you, but generally I try to not overcomplicate a design without a good reason.

[bdunham7]

As high as +-10V.

My plan is to have the external input go into an opamp to scale down the voltage to an expected value, then re-inverted on another opamp voltage follower to get the correct polarity. I need to handle the case where a higher voltage than expected is externally supplied.

I meant how high at the input pin of your IC under normal operation (full-scale) where you would want minimal leakage? 

Why not invert the input first at full (or part) scale and add the protection at the intermediate stage and then do a re-inversion and final scaling with the second op-amp?

[level6]

So use a higher voltage zener to minimize the current loss?

[bdunham7]

So use a higher voltage zener to minimize the current loss?

If it is buffered small differences in current won't matter, but the  main thing is that zeners at a higher voltage will have a sharper knee.  Or you can use more traditional voltage clamp techniques using a diode and the power supplies.

[TimFox]

Low-voltage "Zener diodes" are, in fact Zener diodes and have a soft knee.
"Zener diodes" above roughly 5 or 6 V are, in fact, avalanche diodes that have a much sharper knee.
From what appears to be a cram school:  https://byjus.com/physics/difference-between-zener-breakdown-and-avalanche-breakdown/

Now, a TL431 "adjustable" shunt regulator will have a very sharp knee at 2.5 V, with its cathode (positive terminal) connected to its "reference" pin to form a two-terminal "Zener-like" object.
The lower-voltage TLV431 has a similar sharp knee at 1.24 V, but adding two resistors will increase that voltage.  https://www.onsemi.com/pdf/datasheet/tlv431a-d.pdf

[Johnny B Good]

@TimFox, thank you!

 I was beginning to wonder whether anybody was ever going to mention the "Bleedin' Obvious" TL431 solution. However, I was assuming the +1.9v maximum (according to the datasheet) had been a maximum electrical stress limit rather than a more nebulous maximum gain voltage point.

 The datasheet is unclear on this point (is the chip damaged when going beyond a "2v + 0.7v clamp diode" limit or does it simply mean the gain maxes out somewhere between +1.7 and +1.9 volts?).

 If it's for protection, then the TL431 (or similar) is the right solution but, if it's simply to match the source control voltage limits (0.1 to 4.8 volts perhaps?) to the 0.0 to 2.0v control voltage range, then the resistor divider solution would seem to best meet this requirement.